Question

In $R^{3}$, let $L$ be a straight line passing through the origin. Suppose that all the points on $L$ are at a constant distance from the two planes $P_{1}: x+2 y-z+1=0$ and $P_{2}: 2 x-y+z-1=0$. Let $M$ be the locus of the feet of the perpendiculars drawn from the points on $L$ to the plane $P_{1}$. Which of the following points lie (s) on $M$ ?

• A. $\left(0, \frac{5}{6},-\frac{2}{3}\right)$
• B. $\left(-\frac{1}{6},-\frac{1}{3}, \frac{1}{6}\right)$
• C. $\left(-\frac{5}{6}, 0, \frac{1}{6}\right)$
• D. $\left(-\frac{1}{3}, 0, \frac{2}{3}\right)$

Answer 1

Answer: (B)

$L: \frac{x}{a}=\frac{y}{b}=\frac{z}{c}=\lambda$

$\Rightarrow \frac{a}{I}=\frac{b}{-3}=\frac{c}{-5}$
$\therefore $ Any point on line $L$ is $(\lambda,-3 \lambda,-5 \lambda)$ Equation of line perpendicular to $P_{1}$ drawn from any point on $L$ is
$\frac{x-\lambda}{1}=\frac{y+3 \lambda}{2}=\frac{z+5 \lambda}{-1}=\mu$
$\therefore M(\mu+\lambda, 2 \mu-3 \lambda,-\mu-5 \lambda)$
But M lies on $P _{1}$,
$\therefore \mu+\lambda+4 \mu-6 \lambda+\mu+5 \lambda+1=0$
$ \Rightarrow \mu=\frac{-1}{6}$
$\therefore M\left(\lambda-\frac{1}{6},-3 \lambda \frac{-1}{3},-5 \lambda+\frac{1}{6}\right)$
For locus of $M$,
$x=\lambda-\frac{1}{6}, y=-3 \lambda-\frac{1}{3}, z=5 \lambda+\frac{1}{6} $
$\Rightarrow \frac{x+1 / 6}{1}=\frac{y+1 / 3}{-3}=\frac{z-1 / 6}{-5}=\lambda$
On checking the given point,
we find $\left(\frac{-1}{6}, \frac{-1}{3}, \frac{1}{6}\right)$ satisfy the above eqn.